A Cubic Puzzle


A puzzle site on the internet asked for a solution, in integers, of the equation



The question didnít specify non-zero integers, so we obviously have the trivial solution x = y = z = 0. Even if we stipulate non-zero integers, itís easy to give an infinite family of integer solutions of any equation of the form



simply by setting x = Bw and y = −Aw, in which case the equation reduces to z = s3 where s = x + y + z. Hence we can substitute for z in the above equation, take the cube root of both sides, and re-arrange the terms to give



Now if we set s = (A−B)k for some arbitrary integer k, we can substitute for s and solve for w to give



Consequently for any integer k the cubic (2) is satisfied by the integers x,y,z given by



Of course, these are not the only integer solutions of (2). More generally, any equation of the form (2) can be re-written as



where (again) s = x + y + z. For any chosen value of s (provided it doesnít violate any divisibility requirements) this equation has infinitely many integer solutions given by the Euclidean algorithm. For example, if we choose s = 1000, the original cubic (1) is



This has the infinite family of solutions



More generally, if x0, y0, z0 is any solution of (2), then so is x0 + αj, y0 + βj, z0 + γj provided that



Therefore, setting α = −(B−1), β = (A−1), γ = −(A−B), we have the infinite family of solutions



Since we can find such an infinite family of solution for any chosen value of s (satisfying the divisibility requirements), we have a doubly-infinite set of integer solutions of (1).


The preceding solutions donít require the x and y terms to cancel out, but they typically lead to solutions in which either x or y is negative. By a suitable choice of s we can find solutions with x and y both positive, but then we typically find that the value of z is negative. For example, with s = 495884 the Euclidean algorithm gives the solutions



Another simple approach to generating solutions of (1) is to note that if the integers x0, y0, z0 are a solution then



for any integer j. Consequently, if X, Y, Z are any integers such that (1) is satisfied modulo 16, then the value of j is given by



and from this we can compute the solution



All these methods give solutions that satisfy the original stated puzzle, which asked for integer solutions, but it isnít obvious that any of these methods gives solutions in strictly positive integers.


It is certainly possible to have positive integer solutions to some equations of the form (2). For example, the equation



has the solution x=1, y=2, x=1. Also, the equation



has the solution x=13, y=7, z=8. More in the spirit of the original puzzle, we also have



We also have two solutions for



As an aside, we note that there is also a solution of the ď54321Ē equation with z = 0, so we have a solution of the two-variable relation



In general a two-variable equation of the form



is more tractable, because the left side can be written as A(x+y) − (A−B)y, which implies that x+y must be a divisor of A−B (assuming x and y are co-prime). Therefore, letting s denote x+y and putting this equal to each divisor of A−B, we can compute rational values of x and y by solving the two equations Ax + By = s3 and x + y = s. This gives



In order to make x and y positive, the value of s2 must be between A and B, so we need only check the divisors of A−B that are between √A and √B. This enables us to determine very efficiently all the positive integer solutions (if any) of (5) for any given values and A and B. For example, by this method we can prove that there are no positive integer solutions of



On the other hand, for the equation



we have A Ė B = (22)(6736)(32069), and if we set s = (22)(6736) we get the positive integer solution x = 18795, y = 8153.


Returning to the original three-variable equation, we can adapt the same approach, assuming s and z are given, and computing the values



We know there are no positive integer solutions x,y with z=0, but we can check with z = 1, 2, 3, 4, 5, and 6. (Using equation (4), it is sufficient to check just these six values of z for solutions modulo 16, then compute j, and then shift the values accordingly.) It is not necessary in this case for s to be a divisor of A Ė B, but we can still check the values of s in a suitable range. By this method we find that equation (1) has the positive integer solution x = 27150, y = 1500, z = 1350. A second solution in positive integers is x = 7350, y = 6000, z = 6650. Interestingly, the quantity x Ė y + z equals (2∙3∙5)3 for the first solution and (2∙2∙5)3 for the second.


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